Lời giải:

$x^3-4x^2-12x+27$

$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$

$=x^2(x+3)-7x(x+3)+9(x+3)$

$=(x+3)(x^2-7x+9)$

tu de bai suy ra: 3x(x^3+4x+4)=3x(x+2)^2

3x^3+12x^2+12x=3x(x^2+4x+4)=3x(x+4)^2

cái này dễ mà

= (2x)^3-3(2x)^2*1+2*3x*1^2-1^3

= (2x-1)^3

\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2\)\(=\left[4y-\left(2x+3\right)\right]\left(4y+2x+3\right)=\left(4y-2x-3\right)\left(4y+2x+3\right)\)

bạn đăng ít  thôi dc ko

đăng ít thôi thế này ai làm đc

A,          m²-7m+12=m²-3m+12-4m=m(m-3)-4(m-3)=(m-4)(m-3)

B,           2x⁴-x³+27-54x=x³(2-x)-27(2-x)=(x³-27)(2-x)=(x-3)(x²+3x+9)(2-x)

a) m^2 -7m +12 = m^2 -3m -4m +12

=m(m -3)-4 (m- 3)

=(m-4)(m-3)

b) 2x^4-x^3 -54x+ 27

=(2m^4-x^3)- (54x - 27)

=x^3(2x-1)-27(2x-1)

=(x^3-27)(2x-1)

a,(m-4)(m-3)

b,(x-3)(2x-1)(x^2-3x+9)

\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)

hc tốt nha !!!!!!!!!

KHÔNG BÍT

\(x^4-6x^3+54x-81\)

\(=x^4+3x^3-9x^3+27x^2-27x^2 +81x-27x-81\)

\(=\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27+81\right)\)

\(=x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)

\(=\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)

\(=\left(x+3\right)\left(x-3\right)^3\)

a, = 2 x ² - 6x +9x - 27

=(x-3).2x +9.(x-3)

=(x-3).(2x+9)

b, = 2x²-6xy+xy -3y²

= 2x.(x-3y) +y.(x-3y)

= (x-3y).(2x+y)